Calculating Z Component of Velocity on Earth's Surface | Tips & Examples

[Wizardsblade]

Hey, I was trying doing some calculations and ran across a question I just can’t figure out. Hopefully I explain this well.
I am trying to find the Z component of velocity of an object on earth’s surface after a given time. For example a rock lies on the ground with a tangential velocity of ~450m/s (I believe). 1 second later it will be going v1 speed in the negative z direction and v2 speed in the x direction. I believe v1^2+v2^2=450^2 but I really don't know where to go from here.

 

[Tide]

Define your coordinate system and state the problem a little more carefully. At it stands it is unclear.

 

[Wizardsblade]

At time = 0 the z axis is up and down and the x-axis is forward and backward in the direction of Earth's rotation, i.e. looking at the Earth form above the earth. What I am trying to figure out is the x and z components at any give time. For example at time = 6 hours the x component will be 0 and the z component will be -450m/s, at time = 12 hours z=0 and x=-450m/s.

 

[Wizardsblade]

Wait I think I figured out a way. I did this: t=1s so I figured the number of seconds in a day (24*60*60=86400) and divided 360 degrees by 86400 and got .0041. Then I took the sin of .0041 and multiplied by 450m/s. So I got 7.2x10^-5m/s. Is this correct for the z component?

 

[Tide]

Unless you're standing at the equator you're going to need three basis vectors in the fixed frame of reference to describe those vectors.

What you are calling z corresponds to the radial coordinate in a spherical coordinate system so that a unit vector in the direction is given by

\hat r = \sin \theta \cos \phi \hat i + \sin \theta \sin \phi \hat j + \cos \theta \hat k

With the simple rotation you've defined simply replace \phi with \omega t. Also, what you are calling the x direction corresponds to a vector in the azimuthal direction (\phi) and the corresponding unit vector is given by

\hat \phi = -\sin \phi \hat i + \cos \phi \hat j